The following problem appears in Spivak's Calculus, appendix to chapter 12, "Parametric Representation of Curves"
- Let $x=u(t)$, $y=v(t)$ be the parametric representation of a curve, with $u$ and $v$ differentiable, and let $P=(x_0,y_0)$ be a point on the plane. Prove that if the point $Q=(u(\bar{t}), v(\bar{t}))$ on the curve is closest to $(x_0,y_0)$, and $u'(\bar{t})$ and $v'(\bar{t})$ are not both $0$, then the line from P to Q is perpendicular to the tangent line of the curve at Q (see figure below). The same result holds if Q is furthest from $(x_0,y_0)$.
My question is about the final comment:
The same result holds if Q is furthest from $(x_0,y_0)$.
How do we show this?
Here is a solution to the problem of showing the result for Q being closes to $(x_0,y_0)$
The square of the distance between P and $(u(t), v(t))$ is
$D^2(t)=(u(\bar{t})-x_0)^2+(v(\bar{t})-y_0)^2$
Note that $D^2$ is
- defined for all $t$
- bounded below by $0$
- differentiable (since $u$ and $v$ are)
- has a minimum on any closed interval (because it is continuous)
By assumption, $(u(\bar{t}),v(\bar{t}))$ is the point on the curve closest to $(x_0,y_0)$, and therefore it is the minimum of $D^2$.
Were this assumption absent, we wouldn't be able to conclude that $D^2$ has a minimum in any open interval, correct?
The derivative $(D^2)'(\bar{t})$ must be zero since the minimum there occurs in the interior of any interval containing it.
$$(D^2)'(\bar{t})=2(u(\bar{t})-x_0)u'(\bar{t})+2(v(\bar{t})-y_0)v'(\bar{t})=0\tag{1}$$
One way to view $(1)$ is as a dot product
$$(u'(\bar{t}), v'(\bar{t})) \cdot (u(\bar{t})-x_0, v(\bar{t})-y_0)=0\tag{2}$$
$(u'(\bar{t}), v'(\bar{t}))$ is a vector tangent to the curve at at Q. $(u(\bar{t})-x_0, v(\bar{t})-y_0)$ is the vector from P to Q. If $\theta$ is the angle between these vectors, then $(2)$ tells us that $\cos{\theta}=0 \implies \theta=\frac{\pi}{2}$.
The solution manual reaches this conclusion using a slightly different approach.
Consider two possible cases
Case 1: $u'(\bar{t}) \neq 0$
From $(1)$, we can divide by $u'(\bar{t})$, to obtain
$$\frac{v'(\bar{t})}{u'(\bar{t})}\cdot \frac{y_0-v(\bar{t})}{x_0-u(\bar{t})}=-1$$
$\frac{v'(\bar{t})}{u'(\bar{t})}$ is the slope of the tangent line at $(x,f(x))=(u(\bar{t}), v(\bar{t}))$, and $\frac{y_0-v(\bar{t})}{x_0-u(\bar{t})}$ is the slope of the line that from Q to P.
Case 2: $u'(\bar{t})=0$
By assumption $u'(\bar{t})$ and $v'(\bar{t})$ are not both zero at the same time, so $v'(\bar{t})\neq 0$. Therefore, the tangent vector $(u'(\bar{t}), v'(\bar{t}))$ at $(u(\bar{t}), v(\bar{t}))$ is parallel to the y-axis.
But then from $(1)$ we have
$$(v(\bar{t})-y_0)v'(\bar{t})=0$$
$$\implies (v(\bar{t})-y_0)=0$$
$$\implies (v(\bar{t})=y_0)=0$$
That is, P and Q have the same y-coordinate. The line between them is thus parallel to the x-axis, and perpendicular to the tangent line which is parallel to the y-axis.
Now let me go back to my original question. It seems that what we can say is that if there is a point on the curve the is the most distant from P, then it will have $(D^2)'(t_{max})=0$. Therefore, all the calculations we just went through will be true for that point. Is this why Spivak makes the comment in bold?
I think you've answered your own question, but here's another way to look at it. Say that the max distance is a constant, $M$. Define a new function, say $E^2=M^2-D^2$. then $E^2$ is bounded below by $0$ and so forth, and everything else in either of the two solutions just applies as-is, since the derivative of $M^2$ is zero.
(This is why the sun is directly overhead on the winter solstice at the equator at noon. https://www.weather.gov/cle/seasons)